What does it mean to apply a vector field to a scalar function?

In a textbook I am reading, the author writes:

Be careful not to confuse the notations $fX$ and $Xf$, the first is a smooth vector field, the second is a real valued function.

If we go down into a specific coordinate representation of $X$ it is a function that eats points in a smooth manifold and spits out a vector, so ti would look like this:

$X(p) = (X^1(p), \cdots, X^n(p))$

Similarly we would get:

$fX(p) = f(p)X(p) = f(p)(X(p) = (X^1(p), \cdots, X^n(p)))$

I am having a hard time understanding what $Xf$ means. $f$ spits out scalars, so it;s output cannot be an input of $X$.

The book describes the action of a vector field on a scalar funciton as follows:

$$(Xf)(p) = Xpf$$

Which is not very clear to me, at face value and naively reading the notation, that looks like the scalar product of a vector, which is just $fX$ but we know that it is not what that is supposed to represent.

Solution 1:

The "action" of a vector $X_p$ on a scalar function $f$ is to take the derivative of $f$ in the direction $X_p$. If $f \in C^1(\mathbb{R}^n)$, $p \in \mathbb{R}^n$ and $X_p$ is a vector in $\mathbb{R}^n$, then $X_p f=\frac{d}{dt}(f(p+tX_p))$. More generally, if $f \in C^1(M)$ for some manifold $M$, let $\gamma:(-1,1) \to M$ be any $C^1$ curve such that $\gamma(0)=p$ and $\dot{\gamma}(0)=X_p$. Then $X_p f=\frac{d}{dt}(f(\gamma(t)))|_{t=0}$.

Solution 2:

A smooth vector field $X \colon U \to \mathbb{R}^n$, $U$ open in $\mathbb{R}^n$ can be identified with the partial differential operator $\sum_{j = 1}^{n}X_j(x)\frac{\partial}{\partial x_j}$. Then if $f \in C^{\infty}(U)$, $Xf \in C^{\infty}(U)$ is the function defined by $Xf(x) = \sum_{j = 1}^{n}X_j(x)\frac{\partial f}{\partial x_j}(x)$.